Dear friends,
The lines X(13)X(15) and X(14)X(16) are parallel to the Euler line of a triangle.
Find the following:
1. the coordinates of the orthogonal projections of X(13), X(14), X(15) and X(16) on the Euler line.
2. the distances of the lines X(13)X(15) and X(14)X(16) from the Euler line.
1)
PX13 = ORTHOGONAL PROJECTION OF X(13) ON EULER LINE
= 2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+(a^4+(b^2+c^2)*a^2-2*(b^ 2-c^2)^2)*(a^6-(b^2+c^2)*a^4-( b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^ 4)*(b^2-c^2)) : : (barys)
= on lines: {2,3}, {13, 523}
= [ 0.5098297029399999, -0.3610351635870000, 3.6553058861110000 ]
PX14 = ORTHOGONAL PROJECTION OF X(14) ON EULER LINE
= -2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+(a^4+(b^2+c^2)*a^2-2*(b^ 2-c^2)^2)*(a^6-(b^2+c^2)*a^4-( b^4-3*b^2*c^2+c^4)*a^2+(b^4-c^ 4)*(b^2-c^2)): : (barys)
= on lines {2,3}
= [ 5.9730124263560000, 5.0877355458610000, -2.6383889389300000 ]
PX15 = ORTHOGONAL PROJECTION OF X(15) ON EULER LINE
= 2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+3*(a^6-(b^2+c^2)*a^4-(b^ 4-3*b^2*c^2+c^4)*a^2+(b^4-c^4) *(b^2-c^2))*(c^2-a^2+b^2)*a^2 : : (barys)
= on lines: {2, 3}, {15, 523}, {2452, 11485}, {2453, 11480}
= {X(3), X(1316)}-Harmonic conjugate of PX16
= [ 3.5363285717900000, 2.6574797248290000, 0.1687191777380000 ]
PX16 = ORTHOGONAL PROJECTION OF X(16) ON EULER LINE
= -2*(a^8-(b^2+c^2)*a^6+b^2*c^2* a^4+(b^2-c^2)^2*b^2*c^2)*sqrt( 3)*S+3*(a^6-(b^2+c^2)*a^4-(b^ 4-3*b^2*c^2+c^4)*a^2+(b^4-c^4) *(b^2-c^2))*(c^2-a^2+b^2)*a^2 : : (barys)
= on lines {2,3}
= {X(3), X(1316)}-Harmonic conjugate of PX15
= [ -5.1846308414960000, -6.0404735773240000, 10.2154373468990000 ]
2)
D35 = distance(X(13)X(15), Euler-line) = |(SB-SC)*(SC-SA)*(SA-SB)/(4*S* OH*(SW+sqrt(3)*S))|
D46 = distance(X(14)X(16), Euler-line) = |(SB-SC)*(SC-SA)*(SA-SB)/(4*S* OH*(SW-sqrt(3)*S))|
D35/D46 = cos(ω+π/6)/cos(ω-π/6), where ω = Brocard angle of ABC.
César Lozada
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